3
$\begingroup$

I am trying to formally learn electrodynamics on my own (I only took an introductory course). I have come across the differential form of Gauss's Law.

$$ \nabla \cdot \mathbf E = \frac {\rho}{\epsilon_0}.$$

That's fine and all, but I run into what I believe to be a conceptual misunderstanding when evaluating this for a point charge.

I know the math looks better in spherical coordinates, but I will be using Cartesian.

So when I calculate the divergence I obtain:

$$ \nabla \cdot \mathbf E = \nabla \cdot kQ\langle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\rangle = \frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$

This can further be simplified:

$$\frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3-3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$

Now instinctively I would say that 3-3 is zero and then the while thing is zero everywhere. I am confused as to why (purely mathematically) this expression is not equal to zero at the origin. I completely understand why it physically has to be that way. And I also understand that it is modeled with the delta dirac function. But what (again, mathematically) is stopping me from saying that equation is just zero even at the origin?

$\endgroup$
  • $\begingroup$ Hint: You are applying the rules of differentiation where the field is not defined/singular/not differentiable. $\endgroup$ – Qmechanic Jan 9 '18 at 10:21
  • $\begingroup$ Yes! Very true. Thank you. I should have thought of that. $\endgroup$ – Ben Jan 9 '18 at 11:24
  • 1
    $\begingroup$ It is not customary to use angle brackets to denote components of a vector. Better use () or []. $\endgroup$ – Ján Lalinský Jan 10 at 14:14
  • $\begingroup$ I picked up the angle bracket habit from mathematicians. $\endgroup$ – Ben Feb 19 at 23:11
2
$\begingroup$

What you want to compute is essentially $$\vec\nabla \,\frac{\vec x}{\left|\vec x\right|^3}$$ at the origin. Of course, that doesn't exist as a function since the field is singular. On the other hand, you have already shown that it vanishes everywhere else.

So you need to interpret the expression in a weak sense, i.e. as a distribution, and consider the integral $$\int_{B_\epsilon}\vec\nabla \,\frac{\vec x}{\left|\vec x\right|^3} \,\text{d}^3x$$ over some volume containing the origin, conveniently chosen as a ball of radius $\epsilon$, convert it to a surface integral which does not include the singularity and see that the result is finite.

$\endgroup$
  • $\begingroup$ You used a phrase here that I would like to ask you about. You said "that doesn't exist as a function since the field is singular." I have not studied (much to my dismay) abstract algebra. When mathematicians use the phrase vector FIELD, are they actually talking about a type of field in the AA sense? $\endgroup$ – Ben Jan 11 '18 at 10:15
  • $\begingroup$ @Ben: No, I'm using it in the simple sense of "function defined at every point in space": The electric field assigns a vector to every point $x$, i.e. it is a function of a variable $x\in \mathbb{R}^3$, and in this case, the function is sigular and not differentiable at $x=0$. $\endgroup$ – Toffomat Jan 11 '18 at 12:04
0
$\begingroup$

But what (again, mathematically) is stopping me from saying that equation is just zero even at the origin?

Divergence includes partial derivatives with respect to x, y & z of the respective components of the electric field. Now, we look at the definition of partial derivatives which is often called as the first principle: $$\frac{\partial f(x,y,z)}{\partial x}_{x,y,z = 0} = \lim_{ x\to0} \frac{f(x,0,0)-f(0,0,0)}{x-0}$$ In our case, if we take the partial derivative of, say x component of the electric field, with respect to x then we would use the definition, but $f(0)$ or the x component of electric field at the origin is undefined. It is called as a singularity. And still, we want to respect gauss's theorem of divergence which is kind of very elegant & intuitive. So, to find a way out of it, we physicists (actually, Paul Dirac) came up with Delta functions (to which mathematicians are unhappy).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.